Diffusion equation with time-dependent boundary condition 
I was trying to solve this 1D diffusion problem
  \begin{equation}
\dfrac{\partial^2 T}{\partial \xi^2} = \dfrac{1}{\kappa_S}\dfrac{\partial T}{\partial t}\, , \label{eq_diff_xi}
\end{equation}
  with the boundary conditions
  \begin{align}
&T(\xi = 2Bt^{1/2},t) = A t^{1/2}\, ,\\
&T(\xi=\infty,t) = 0\, ,\\
&T(\xi,0) = 0\, ,
\end{align}
  where $A$ is a constant.

I know that the solution is $T = A \sqrt{t}\, \rm{erfc}(\xi/2\sqrt{kt})/\rm{erfc}(B\sqrt{k})$
I tried by using the Laplace transformation, but I found a problem since I have conditions on $\xi = 2Bt^{1/2}$ instead of $\xi = 0$.
More precisely, if the Laplace function of $T(\xi,t)$ is $\Theta(\xi,s)$, after apply the Laplace transformation plus $T(\xi=\infty,t) = 0$ and $T(\xi,0) = 0$, I got 
\begin{equation}
\Theta(\xi,s) = C_1(s)\exp{\left(-\sqrt{\dfrac{s}{\kappa_T}}\xi\right)}\, .
\end{equation} 
So now, to find $C_1(s)$ and use the convolution property of the Laplace transformation, I need a condition on $\xi = 0$, but I only now that 
$T(\xi = 2Bt^{1/2},t) = A t^{1/2}$. 
Does any of you know if the Laplace transform has some other properties that allow me to solve the problem?
 A: The boundary condition hints to try a change of variables. Let's will be looking for the solution in the form 
$$
T(\xi,t) = A\sqrt{t}\ \tau(\xi/2B\sqrt{t},t)
$$
Then boundary conditions and equation for $\tau(x,t)$ are respectivly
$$
\tau(1,t) = 1,\qquad \tau(\infty,t) = 0
$$
and
$$
t\frac{\partial\tau}{\partial t}(x,t) = \frac12\left(x\frac{\partial\tau}{\partial x}(x,t) - \tau(x,t)\right)+\frac{k}{4B^2}\frac{\partial^2\tau}{\partial x^2}(x,t)
$$
This problem is solvable by the separation of varfiables method. If we will look for the solution in the form $\tau(x,t) = f(x)g(t)$, then we'll get
$$
t\frac{\dot{g}(t)}{g(t)} = \frac1{2f(x)}\left(xf'(x)-f(x)+\frac{k}{2B^2}f''(x)\right) = \lambda = const
$$
Boundary condition looks like $\tau(1,t) = f(1)g(t) = 1$. It follows that $\lambda = 0$. Then the equation for $f(x)$ is
$$
\frac{k}{2B^2}f''(x)+xf'(x)-f(x) = 0.
$$
If we choose $g(t) = 1$, then boundary conditions for $f(x)$ are
$$
f(1) = 1,\qquad f(\infty) = 0.
$$
It's up to you to check if a solution to this problem gives the known solution.
